Another way appears on a comment on your [question](https://math.stackexchange.com/questions/3561826/proof-of-int-mathbfr-frac-mathrmdx-gamma-alpha-x-gamma-beta), so this is just a rip-off trying to make things tidier but surely there are other ways to this and Titchmarsh  to prove it

$
\int_{\mathbb{R}}\frac{dx}{\Gamma(\alpha+x)\Gamma(\beta-x)}\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\frac{\Gamma(\alpha+\beta-1)dx}{\Gamma(\alpha+x)\Gamma(\beta-x)}\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\frac{(\alpha+\beta-2)!dx}{(\alpha+x-1)!(\beta-x-1)!}\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\binom{\alpha+\beta-2}{\alpha+x-1}dx\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-1}}{z^{\alpha+x}}dzdx\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-1}}{z^{\alpha}}\int_{\mathbb{R}}z^{-x}dxdz\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-1}}{e^{i\alpha}}\int_{\mathbb{R}}e^{-i\theta}dxd\theta\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-1}}{e^{i\alpha}}\delta(-\theta)d\theta\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-1}}{e^{i\alpha\theta}}\delta(\theta)d\theta\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\frac{(1+e^{i0})^{\alpha+\beta-1}}{e^{i\alpha 0}}\\
=\frac{2^{\alpha+\beta-1}}{\Gamma(\alpha+\beta-1)}
$